I Determining H value from scalefactor

[JohnnyGui]

I’m probably missing something obvious here but I can’t figure out what.

Consider a galaxy, starting at an initial distance of $D_0$, recessing with a constant velocity. After a time Δt, we’d measure a larger distance $D_{t}$. From this scenario I’d conclude that the Hubble value based on this object is calculated by:
$$H = \frac{D_t – D_0}{Δt} \cdot \frac{1}{D_0}$$
That is, the velocity, which is obtained by the first fraction, divided by the initial distance $D_0$ of the galaxy gives the Hubble value when the galaxy was at distance $D_0$

Now, from what I understand, the scale factor $a$ is how large the reached distance $D_t$ is relative to the initial distance $D_0$. Thus; $a = \frac{D_t}{D_0}$
Furthermore, the rate of change of the scale factor $\dot a$ is how much the distance has increased per unit of time, relative to the initial distance $D_0$. So $\dot a$ could be calculated by adding the velocity (a distance per unit time) to the initial distance $D_0$ and dividing that by $D_0$:
$$\dot a = (\frac{D_t – D_0}{Δt} + D_0) \cdot \frac{1}{D_0}$$
When combining these formulations for $H$, $a$ and $\dot a$ I’d get: $H+1 = \dot a$.
This is obviously not true since it should be $H = \frac{\dot a}{a}$.

What am I missing?

 

[PeterDonis]

from what I understand, the scale factor aa is how large the reached distance $D_t$ is relative to the initial distance $D_0$.

Not with the definition of $H$ you are using. If $H = \dot{a} / a$, then $a$ has units of distance, which means $a$ at time $t$ is just $D_t$, and $a$ at time $t_0$ is just $D_0$.

So $\dot a$ could be calculated by adding the velocity (a distance per unit time) to the initial distance $D_0$

No. The derivative $\dot{a}$ is just the change in distance over the change in time (in the limit as the change in time goes to zero), so it would be

$$
\dot{a} = \lim_{\Delta t \rightarrow 0} \frac{D_t - D_0}{\Delta t}
$$

So $H = \dot{a} / a$ at time $t_0$ would be

$$
\frac{\dot{a}}{a} = \frac{1}{D_0} \lim_{\Delta t \rightarrow 0} \frac{D_t - D_0}{\Delta t}
$$

 

[John Park]

So ˙ a a could be calculated by adding the velocity (a distance per unit time) to the initial distance D 0 D 0 and dividing that by D 0 D 0 :

Your resulting equation for a-dot makes no sense dimensionally.

 

[JohnnyGui]

Not with the definition of HHH you are using. If H=˙a/aH=a˙/aH = \dot{a} / a, then aaa has units of distance, which means aaa at time ttt is just DtDtD_t, and aaa at time t0t0t_0 is just D0D0D_0.

Thanks for the reply. Could you please explain a bit more why my defintion of $H$ says that $a = D_0$ or $a = D_t$ at $t_0$ and $t$ respectively? I thought a scale factor is the ratio between the new $D_t$ and the original distance $D_0$.

 

[PeterDonis]

I thought a scale factor is the ratio between the new $D_t$ and the original distance $D_0$.

It isn't. It's just the distance. More precisely, it's just the distance if you are defining $H$ as $\dot{a} / a$. That should be evident from the fact that $H$ is the fractional rate of change of distances, i.e., the ratio of recession speed to distance, which is the ratio of distance per unit time to distance. So if $H = \dot{a} / a$, then $\dot{a}$ must be recession speed--distance per unit time--and $a$ must be distance.

 

[JohnnyGui]

It isn't. It's just the distance. More precisely, it's just the distance if you are defining $H$ as $\dot{a} / a$. That should be evident from the fact that $H$ is the fractional rate of change of distances, i.e., the ratio of recession speed to distance, which is the ratio of distance per unit time to distance. So if $H = \dot{a} / a$, then $\dot{a}$ must be recession speed--distance per unit time--and $a$ must be distance.

I'm surprised, because this wiki does use $D_t = a \cdot D_0$ and $H = \frac{\dot a}{a}$ simultaneously. From the article it does look like they distinguish the distance from the scale factor.

 

[PeterDonis]

From the article it does look like they distinguish the distance from the scale factor.

What they are doing amounts to measuring distances in units of $D_0$ instead of in units of, say, light-years. So $a = 1$ just means the distance is $D_0$ in units of light-years (or whatever ordinary units you want to use).

Mathematically, if you really want to interpret $D_0$ this way, as a distance unit, then you would write

$$
H = \frac{\dot{a}}{a} = \frac{D_0}{D_t} \frac{d}{dt} \frac{D_t}{D_0}
$$

Since $D_0$ doesn't change with time, the $D_0$ factors just cancel out and you have the same thing I wrote before. In terms of the limits, you would have

$$
\dot{a} = \lim_{\Delta t \rightarrow 0} \frac{a_t - a_0}{\Delta t} = \frac{1}{D_0} \lim_{\Delta t \rightarrow 0} \frac{D_t - D_0}{\Delta t}
$$

And the $D_0$ will just cancel out, once again, when you evaluate $\dot{a} / a$.

 

[JohnnyGui]

Explanation

Thanks a lot for the detailed explanation. It makes sense now.

I just noticed my error myself as well. The Wiki is considering $D_t$ as a distance in the past while $D_0$ being the current distance at this moment. I was considering them to be the other way around whch led me to erroneously think that $a$ should be >1.

After correcting for this, when one writes the distance in units of $D_0$, the $\dot{a}(t)$ is the speed at distance $D_t$ relative to the distance $D_0$ just like you wrote it.
If the speed has been constant all throughout history, then $\dot{a}(t)$ would equal the current Hubble value $H_0$. If the speed hasn't been constant, then that means one would have determine $\dot{a}(t)$ at time $t$ and then use...
$$\dot{a}(t) \cdot D_0 = v(t)$$
...to calculate the speed at time $t$. This calculated speed divided by the distance $D_t$ at that time would give the Hubble value at that time, $H(t)$.
$$\frac{\dot{a}(t) \cdot D_0}{D_t} = H(t)$$
Since $D_t = a(t) \cdot D_0$ (a(t) being <1) this means that:
$$\frac{\dot{a}(t) \cdot D_0}{a(t) \cdot D_0} = \frac{\dot{a}(t)}{a(t)} = H(t)$$